Question

\[\huge{\int\limits\limits\limits\limits_{0}^{{\frac \pi 2}}}\int\limits\limits\limits\limits_{0}^{\cos \theta} e^{\sin \theta}d r d \theta\]

Answer 1

If you integrate first with respect to r, you will be able to use a nice substitution afterwards.

\[ u= sin \theta\]

Answer 2

\[{\int\limits_{0}^{{\frac \pi 2}}}\int\limits\limits\limits\limits_{0}^{\cos \theta} e^{\sin \theta}d r d \theta\]\[={\int\limits_{0}^{{\frac \pi 2}}}e^{\sin \theta}\int\limits_{0}^{\cos \theta} d r d \theta\]

Answer 3

uh, can you put "huge" in front?

Answer 4

it makes the text larger

Answer 5

\[=\huge{\int\limits_{0}^{{\frac \pi 2}}}e^{\sin \theta}\int\limits_{0}^{\cos \theta} d r d \theta\]

Answer 6

\[=\huge{\int\limits_{0}^{{\frac \pi 2}}}e^{\sin \theta}\left(r|_{0}^{\cos \theta}\right) d \theta\]

Answer 7

I get A=e-1

Answer 8

\[=\huge{\int\limits_{0}^{{\frac \pi 2}}}e^{\sin \theta}{\cos \theta} d \theta\]

Answer 9

Um, this equation looks like it's missing an r. usually double integrals of this type are integrated with respect to r*dr*d(theta)

Answer 10

that is a good point @agentc0re

Answer 11

The "r" wasn't in the question. It's question 5.

Answer 12

Guess it's been left away so the substitution becomes more simple.

Answer 13

And I don't even know what was done...

Answer 14

I'm starting to despise double integrals and I just started.

Answer 15

lol. just wait until you do triple integrals and have to change your bounds of integration. :D Honestly it's not too bad. My guess is that it's not a true volume oriented problem so there is no r*dr*d(theta).

Looks like you were doing good before though, keep it up!

Answer 16

\[{\int\limits_{0}^{{\frac \pi 2}}}e^{\sin \theta}{\cos \theta} d \theta\]
\[\text{let } u=\sin \theta;\qquad\text du=\cos\theta \text d\theta\]
\[\theta =\frac{\pi}{2}\longrightarrow u=1\]\[\theta =0\longrightarrow u=0\]
\[={\int\limits_{0}^{1}}e^{u}\text du\]

Answer 17

@agentc0re uh huh, right. Keep up with the sarcasm and I'll find an Urk-Khai!

Answer 18

can you finish from here/?

Answer 19

So the answer would be e-1?

Answer 20

yes @roadjester